Question

如果b分之a等于6+ab均为非零数非零自然数那么ab的最大公因数是多少最小公倍数

Answer 1

问：如果b分之a等于6+ab均为非零数非零自然数那么ab的最大公因数是多少最小公倍数

答：根据题意，我们有以下等式：b/a = 6 + ab移项得：b = a(6 + ab)由于a和b是两个正整数，所以6 + ab > a，因此有ab > 6。因此可能的组合为：ab = 7, a = 1, b = 7 或 a = 7, b = 1ab = 8, a = 1, b = 8 或 a = 2, b = 4ab = 9, a = 1, b = 9 或 a = 3, b = 3ab = 10, a = 1, b = 10 或 a = 2, b = 5ab = 11, a = 1, b = 11ab = 12, a = 1, b = 12 或 a = 3, b = 4ab = 14, a = 1, b = 14 或 a = 2, b = 7ab = 15, a = 1, b = 15 或 a = 3, b = 5ab = 16, a = 1, b = 16 或 a = 2, b = 8ab = 17, a = 1, b = 17ab = 18, a = 1, b = 18 或 a = 2, b = 9 或 a = 3, b = 6ab = 19, a = 1, b = 19ab = 20, a = 1, b = 20 或 a = 2, b = 10 或 a = 4, b = 5ab = 21, a = 1, b = 21 或 a = 3, b = 7ab = 22, a = 1, b = 22 或 a = 2, b = 11ab = 24, a = 1, b = 24 或 a = 3, b = 8 或 a = 4, b = 6根据辗转相除法，可以求出所有最大公因数和最小公倍数的值：(1, 7): GCD=1, LCM=7(1, 8): GCD=1, LCM=8(1, 9): GCD=1, LCM=9(1, 10): GCD=1, LCM=10(1, 11): GCD=1, LCM=11(1, 12): GCD=1, LCM=12(1, 14): GCD=1, LCM=14(1, 15): GCD=1, LCM=15(1, 16): GCD=1, LCM=16(1, 17): GCD=1, LCM=17(1, 18): GCD=1, LCM=18(1, 19): GCD=1, LCM=19(1, 20): GCD=1, LCM=20(1, 21): GCD=1, LC